# Questions tagged [lie-superalgebras]

The lie-superalgebras tag has no usage guidance.

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### super Lyndon words

Lyndon words form a basis for Free Lie algebras.
In this direction, I need the reference for the super Lyndon words for free Lie superalgebras.
Given the definition of super Lyndon words, how to ...

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### Lyndon basis of free Lie superalgebras

Lyndon basis for Free Lie algebras is well known in the literature.
My question is,
what is the analogous combinatorial model for the case of free Lie superalgebras? what is the super analogous of ...

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### Reference Request: Representation Theory of Real Lie Superalgebras

Are there some references for the representation theory real lie superalgebras, specifically of $psu(1,1|2)$, and $u(1,1|2)$?

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### Serre relations for Lie Superalgebras

Finite dimensional complex simple Lie algebras are classified using Cartan matrices. One of the main ingredients is Serre Relations. Lets call this Cartan-Killing theory.
I have the following ...

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### Coordinate ring of an equivariant embedding of a homogeneous projective variety

Lie algebra: Let $G$ be a semisimple, simply connected linear algebraic group with a fixed Borel subgroup $B$. Let $P$ be a parabolic subgroup containing $B$. Let $\lambda$ be a dominant integral ...

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### Kazhdan-Lusztig equivalence for Lie super-algebras

Let $\mathfrak g$ be a semi-simple Lie algebra. Kazhdan and Lusztig studied the category of representations of the corresponding affine Lie algebra (the central extension of $\mathfrak g((t))$) which ...

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### odd positive roots of basic classical Lie superalgebras

In the Lie algebra case, positive roots are "almost ($S_{\alpha}$ permutes $\Delta^+ -\{\alpha\}$)" invariant under simple reflections. A similar statement I want to understand for Lie superalgebras.
...

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### action of Weyl group element on Weyl vector

Let $\mathfrak g = \mathfrak g_0 \oplus \mathfrak g_1$ be a basic classical Lie super algebra and let $\rho = \text{half sum of even positive roots} - \text{half sum of odd positive roots}$ be the ...

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### finite dimensional modules are highest weight modules [closed]

Let $\mathfrak{g}$ be a basic classical simple Lie super algebra. I want to prove that every finite dimensional module over $\mathfrak{g}$ has a highest weight vector.
My feeling is, since $e_i$'s ...

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### how the borel subalgebras of p(n) look like except the standard one?

I am reading the book Musson: Lie superalgebras. In Chapter 3, on page 62, Lemma 3.6.8 tells about support of Borel subalgebra. I am confused about this.
I am trying to do the proof of the Proposition ...

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### Character formula for Lie superalgebras

The Weyl character formula and the denominator identity play important roles in the representation theory of classical simple Lie algebras and Kac-Moody Lie algebras over $\mathbb{C}$
Can you suggest ...

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### $P(1)$ strange type classical Lie superalgebras

I am reading the book "Lie superalgebras and enveloping algebras" by Ian M. Musson.
The strange type $P(n)$ series of Lie superalgebras are defined (§2.4.1, p. 17) only for $n \ge 2$ even though for $...

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### left ideals in Lie super algebras

Let $\mathfrak g$ be a Lie superalgebra.
If $\mathfrak a$ is not a grade subspace of $\mathfrak g$, then why does $[\mathfrak g, \mathfrak a]$ and $[\mathfrak a, \mathfrak g]$ are not same?
For me ...

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286 views

### Graph of a Lie super algebra

Let $A$ be a generalized Cartan matrix and let $\mathfrak{g}$ be the Kac-Moody Lie algebra associated to $A$. There is an associated graph of $\mathfrak{g}$ which is known as the Dynkin diagram of $\...

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### Two definitions of super-Virasoro algebra

Let $A=\mathbb C[x,\epsilon]$ where $x$ is an even variable and $\epsilon$ is an odd variable (thus $A$ is a commutative super-algebra). Let $\mathfrak g$ denote the Lie super-algebra of vector fields ...

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### Vanishing of H-cohomology

This looks elementary, but somehow I am stuck, please bear with me:
Let $H$ be a differential 3-form, nowhere vanishing, but not necessarily closed. What is a sufficient condition that the sequence ...

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### Dual Coxeter Number for Superalgberas

I am looking for a reference that gives the definition and has summarized the dual Coxeter number for superalgebras, especially for $\mathfrak{u}(m|n)$ (the Lie algebra of unitary supergroup $U(m|n)$)....

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### Action of orthosymplectic group $SpO(d+1|d)$ on $PGL(d+1|d)$

Suppose $d$ is odd, and consider the super vector space $\mathbb{C}^{d+1|d}$ and its super projectivization $\mathbb{P}^{d|d}$.
In the case $d=1$, a paper of Witten's, concerning the moduli space of ...

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### Noncommutative cohomology of flag varieties

Consider the Grassmannian $Gr(n, N)$ of $n$-dimensional subspaces of $\mathbf C^N$. Its cohomology ring is isomorphic to $\mathbf C[x_1, \ldots, x_n, \bar x_1, \ldots \bar x_{N-n}]/I_{n,N}$, wherethe ...

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### Endomorphismensatz for Lie superalgebras

For semisimple complex Lie algebras there is Soergel's Endomorphismensatz
$$C = \operatorname{End}(P(w_0)) \cong \mathbf C[\mathfrak h]/\mathbf C[\mathfrak h]^W$$
for $w_0$ the longest element in ...

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### Is there a name for Lie superalgebras which are generated by the odd subspace?

Every Lie superalgebra $\mathfrak{g} = \mathfrak{g}_{\bar 0} \oplus \mathfrak{g}_{\bar 1}$ has a canonical ideal $\mathfrak{k} = [\mathfrak{g}_{\bar 1}, \mathfrak{g}_{\bar 1}] \oplus \mathfrak{g}_{\...

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### Definition of orthosymplectic supergroups

I found two versions of definitions of orthosymplectic supergroups. It seems that they are not equivalent. I don't know which version of the definition is standard.
The first version of the ...

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### Reference request: coordinate ring of $OSP(2p|n)$

In the paper, the orthosymplectic supergroup $OSP(2p|n)$ is defined as follows.
Let $A = A_0 \oplus A_1$ be a supercommutative superalgebra, where elements in $A_0$ are even and elements in $A_1$ are ...

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### Super version of Poisson brackets of tensor products

Let $A$ be a Poisson super algebra ($A$ is a super algebra and $A$ satisfies super Jacobi identity, super commutativity, super Leibniz rule).
Super version of the product of two tensor products is
\...

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### Super-extensions of the Poincaré Lie algebra

For $(\mathfrak{g},[-,-])$ an ordinary Lie algebra let me say that a super-extension of it (maybe not the best terminology) is a super-Lie algebra $(\mathfrak{s}, [-,-]_{\mathfrak{s}})$ whose bosonic ...

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### Classical Yang-Baxter equation for Lie algebras and Lie superalgebras

The classical Yang-Baxter equation is
\begin{align}
[r_{12}, r_{13}] + [r_{12}, r_{23}] + [r_{13}, r_{23}] = 0. \quad (1)
\end{align}
What are the differences between this equation in the case of Lie ...

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### References request: vector representations of Lie superalgebras

Are there some references of fundamental representations of Lie superalgebras (in particular for the Lie superalgebra $sl(m|n)$? Thank you very much.

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### Two definitions of the super Jacobi identity

In this paper, page 149, the super Jacobi identity is given by
\begin{align}
J(x, y,z) := (-1)^{|x||z|}[[x, y],z] +(-1)^{|z||y|}[[z,x], y]+(-1)^{|y||x|}[[y,z],x] = 0.
\end{align}
But in this article, ...

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### Solution of the Yang-Baxter equation associated to the $U_q[osp(2n+2|2m)^{(2)}]$ Lie superalgebra

I have a solution (a $R$ matrix) of the Yang-Baxter equation,
\begin{equation}
R_{12}(x_{1})R_{13}(x_{1}x_{2})R_{23}(x_{2})=R_{23}(x_{2})R_{13}(x_{1}x_{2})R_{12}(x_{1})
\end{equation}
that probably ...

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### Irreducible representations in BGG category $\mathcal{O}$ over (finitely) direct sum of general linear Lie superalgebra

Let $\mathfrak{g} = \oplus_i^k\mathfrak{gl}(m_i|n_i)$ be a direct sum of general linear Lie superalgebras $\mathfrak{gl}(m_i|n_i)$'s with the Cartan subalgebra $\mathfrak{h} = \oplus_i^k \mathfrak{h}...

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### Mirror Symmetry for Flag Supermanifolds

I recently learned the following two things, and I wish to know how to make them reconciled.
(1) As far as I understand, the flag manifolds serve as a tractable class of examples for the very ...

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### Why, in terms of quantum groups, does the knot determinant appear as an evaluation of both the Jones and Alexander polynomials?

The Jones polynomial can be computed from the representation theory of $\mathcal{U}_q(\mathfrak{sl}(2))$. The Alexander polynomial has an analogous description in terms of the representation theory of ...

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### Non-graded representations over Lie superalgebra $\mathfrak{gl}(m,n)$

I have the following questions:
Let $m,n$ be positive integers. Consider representations over the general linear Lie super-algebra $\mathfrak{gl}(m,n)$. Namely, modules over the associative algebra $U(...

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### Orthosymplectic group, matrix representations

We have the orthosymplectic $osp(n,m|2k)$. The bosonic part is $so(n,m)\times sp(2k)$. The lie algebra generators are given in eg
http://cds.cern.ch/record/524737/files/0110257.pdf$
where the group ...

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### homomorphism of Lie superalgebras

In the book Shun-Jen Cheng, Weiqiang Wang Dualities and Representations of Lie Superalgebrasm. One founds the following definition(Definition 1.3):
Let $\mathfrak{g}$ and $\mathfrak{g'}$ be Lie ...

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### Kauffman's state model for the Alexander polynomial, via representation theory

I've been reading Oleg Viro's paper on "quantum relatives of the Alexander polynomial" (arXiv:math/0204290), which, among other more general things, derives state-sum formulas for the Alexander ...

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### Linear independence in (graded) Lie algebras

I asked a mixed-up version of this question earlier.
The Lie algebras I have in mind are the homotopy Lie algebras of wedges of finitely many spheres (in dimensions greater than $1$). Thus each ...

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### Definition of the supertrace in superalgebra representations

Let us consider a matrix superalgebra $A$ with generators satisfying $[L_a,L_b]=i L_c f^c{}_{ab}.$ The generators are matrices on which supertrace is defined bu the usual trace on the bosonic part ...

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### What are the simple Lie superalgebras of type E?

Background
Simple finite dimensional Lie superalgebras over $\Bbb C$ have been classified. There are the Cartan type superalgebras (algebras of purely odd vector fields), two strange families P(n) ...

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### Lie superalgebra in two dimensions

The standard formulation of two dimensional $N=(2,2)$ and $N=(0,2)$ supersymmetry algebras in physics is an explicit one; I am not aware of the underlying abstract Lie superalgebras. Does anyone know ...

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### Finite dimensional consistently graded Lie superalgebras of depth greater than 2

Victor Kac, in the paper
"Classification of infinite-dimensional simple linearly compact Lie superalgebras", http://www.mat.univie.ac.at/~esiprpr/esi605.pdf
writes at the beginning of section 5 (p....

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### Is the SUSY Algebra isomorphic for all Kähler Manifolds?

For a Kähler manifold, the graded algebra generated by $\partial,\overline{\partial},\partial^*,\overline{\partial}^\ast$, the Lefschetz operator, and the dual Lefschetz operator, is called the ...

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### Do all finite $W$-superalgebras have 1-dimensional representations?

Premet proved the famous KW-conjecture in modular Lie algebra.
After, Premet introduced the finite $W$-algebra $U(g, e)$.
Also, Premet proposed the conjecture every algebra $U(g, e)$ admits a $1$-...

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### I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt

Question: I am talking about the proof given on pages 50-52 of Pierre Deligne, Pavel Etingof, Daniel S. Freed, Lisa C. Jeffrey, David Kazhdan, John W. Morgan, David R. Morrison, and Edward Witten (...

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### Building Lie-like algebras from modules over semisimple Lie algebras

Here is a construction of a very broad class of "Lie-like" algebras, and I want to know more about them.
Here is the main definition: Suppose $\mathfrak{g}$ is a complex semsimple Lie algebra and $\...

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### Is there much theory of superalgebras acting on manifolds by alternating polyvector fields?

Usual story: vector fields on $M$, with their Lie bracket, form a Lie algebra. We can consider "actions" of some other Lie algebra ${\mathfrak g}$ on $M$ by looking at Lie homomorphisms ${\mathfrak g}\...

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### Is there a definition of analogue Weyl group for Lie super algebra?

I heard from some people working in Lie super algebra that there was no proper definition for Weyl group of Lie super algebra. I do not know Lie super algebra at all. But When I searched on Google, I ...